University Quantum Mechanics II · unit UQMII-14
The Adiabatic Approximation and Berry's Phase
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
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Adiabatic invariants: a pendulum of slowly changing length: infer an input with uncertainty
An adiabatic invariant is a quantity that stays fixed when a parameter of a periodic system changes slowly compared with one period, even though the energy changes; for an oscillator it is E/ω, the area of its phase-space orbit divided by 2π. Use the supplied topic model Y_out = (r^1.5+r^(-1.5))/2, where Y_out is MEAN SUDDEN I₂/I₁ OVER φ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Length ratio L₂/L₁, dimensionless. Keep phi (Phase φ of the sudden pull) = 60 ° fixed. For this exercise use only 0.553125 ≤ r ≤ 0.946875 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 1.0609198, with a bounded reading uncertainty ±0.0086348. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000039375, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
MEAN SUDDEN I₂/I₁ OVER φ against r
Predictions from the supplied Adiabatic invariants: a pendulum of slowly changing length model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): MEAN SUDDEN I₂/I₁ OVER φ
Horizontal axis (x): r: Length ratio L₂/L₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Length ratio L₂/L₁ | MEAN SUDDEN I₂/I₁ OVER φ |
|---|---|
| 0.553125 | 1.4211304273382364 |
| 0.6515624999999999 | 1.2136520792317815 |
| 0.75 | 1.0945598853386656 |
| 0.8484375 | 1.0305448061550542 |
| 0.9468749999999999 | 1.0033542277985958 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(r) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.553125) = 1.42113 and Y_out(0.946875) = 1.00335. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 0.553125 + (1.06092 − 1.42113)(0.946875 − 0.553125)/(1.00335 − 1.42113) = 0.89262. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives r = 0.793312; substituting back gives Y_out = 1.0609198.
- The interpolation error in the input is |r_line − r| = 0.0993075. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.052285 and 1.0695546. Sort the roots even if the response decreases: 0.780961 ≤ r ≤ 0.806821.
- The lower and upper input deviations are 0.0123513 and 0.0135085. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer r ≈ 0.793312; compatible interval [0.780961, 0.806821]. Endpoint interpolation gives 0.89262.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The adiabatic theorem and its validity condition: infer an input with uncertainty
The adiabatic theorem says that a system starting in a non-degenerate eigenstate of a slowly changing Hamiltonian ends in the matching eigenstate |n(t)⟩ of the Hamiltonian at that moment, changed only by a phase, where slowly means ħ|⟨m|Ḣ|n⟩| ≪ (Eₙ − Eₘ)² for every other state m. Use the supplied topic model Y_out = (fr/f0)^2, where Y_out is PEAK, FIRST ORDER (Ω/ω₀)² (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. f0 is Gap frequency ω₀/2π, measured in MHz. Keep fr (Turning rate Ω/2π) = 0.5 MHz fixed. For this exercise use only 1.31875 MHz ≤ f0 ≤ 3.68125 MHz on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.032821708, with a bounded reading uncertainty ±0.0021561. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625 MHz, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
PEAK, FIRST ORDER (Ω/ω₀)² against f0
Predictions from the supplied The adiabatic theorem and its validity condition model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): PEAK, FIRST ORDER (Ω/ω₀)²
Horizontal axis (x): f0: Gap frequency ω₀/2π / MHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f0: Gap frequency ω₀/2π / MHz | PEAK, FIRST ORDER (Ω/ω₀)² |
|---|---|
| 1.31875 | 0.14375238651422922 |
| 1.909375 | 0.0685736939523895 |
| 2.5 | 0.04000000000000001 |
| 3.090625 | 0.026172630993506935 |
| 3.68125 | 0.018448004012440875 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(f0) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.31875) = 0.143752 and Y_out(3.68125) = 0.018448. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives f0_line = 1.31875 + (0.0328217 − 0.143752)(3.68125 − 1.31875)/(0.018448 − 0.143752) = 3.41025 MHz. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives f0 = 2.75988 MHz; substituting back gives Y_out = 0.032821708.
- The interpolation error in the input is |f0_line − f0| = 0.650372 MHz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.030665608 and 0.034977808. Sort the roots even if the response decreases: 2.67346 MHz ≤ f0 ≤ 2.85525 MHz.
- The lower and upper input deviations are 0.0864149 MHz and 0.0953755 MHz. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer f0 ≈ 2.75988 MHz; compatible interval [2.67346, 2.85525] MHz. Endpoint interpolation gives 3.41025 MHz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The sudden approximation, the opposite limit: infer an input with uncertainty
The sudden approximation treats a Hamiltonian switched in a time short compared with ħ over the relevant level spacings as switched instantly: the state is unchanged, and each new eigenstate |φₙ⟩ is found with probability |⟨φₙ|ψ⟩|². Use the supplied topic model Y_out = n^2/r^2, where Y_out is NEW LEVEL E′ₙ / E₁ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is New width L′/L, dimensionless. Keep n (New level shown, n) = 1 fixed. For this exercise use only 1.31875 ≤ r ≤ 3.68125 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.13128683, with a bounded reading uncertainty ±0.0086242. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
NEW LEVEL E′ₙ / E₁ against r
Predictions from the supplied The sudden approximation, the opposite limit model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): NEW LEVEL E′ₙ / E₁
Horizontal axis (x): r: New width L′/L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: New width L′/L | NEW LEVEL E′ₙ / E₁ |
|---|---|
| 1.31875 | 0.5750095460569169 |
| 1.909375 | 0.27429477580955797 |
| 2.5 | 0.16 |
| 3.090625 | 0.10469052397402774 |
| 3.68125 | 0.07379201604976349 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(r) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.31875) = 0.57501 and Y_out(3.68125) = 0.073792. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 1.31875 + (0.131287 − 0.57501)(3.68125 − 1.31875)/(0.073792 − 0.57501) = 3.41025. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives r = 2.75988; substituting back gives Y_out = 0.13128683.
- The interpolation error in the input is |r_line − r| = 0.650372. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.12266263 and 0.13991103. Sort the roots even if the response decreases: 2.67346 ≤ r ≤ 2.85525.
- The lower and upper input deviations are 0.086413 and 0.0953731. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer r ≈ 2.75988; compatible interval [2.67346, 2.85525]. Endpoint interpolation gives 3.41025.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Landau-Zener transitions at a swept avoided crossing: infer an input with uncertainty
A Landau-Zener transition is the jump of a two-level system across an avoided crossing whose diabatic energies are swept past each other at a constant rate α; for coupling V it happens with probability exp(−2π|V|²/(ħα)), and otherwise the system follows the adiabatic curve into the other state. Use the supplied topic model Y_out = exp(-2*pi*v^2/(0.6582*al)), where Y_out is JUMP PROBABILITY P_D (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. v is Coupling |V|, measured in μeV. Keep al (Sweep rate α) = 2 μeV ns⁻¹ fixed. For this exercise use only 0.34875 μeV ≤ v ≤ 1.45125 μeV on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.0068860479, with a bounded reading uncertainty ±0.0010264. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00011025 μeV, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
JUMP PROBABILITY P_D against v
Predictions from the supplied Landau-Zener transitions at a swept avoided crossing model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): JUMP PROBABILITY P_D
Horizontal axis (x): v: Coupling |V| / μeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Coupling |V| / μeV | JUMP PROBABILITY P_D |
|---|---|
| 0.34875 | 0.5596048709637621 |
| 0.624375 | 0.15555926662931807 |
| 0.9 | 0.020939143258989837 |
| 1.1756250000000001 | 0.001364804826296571 |
| 1.45125 | 0.00004307554800144156 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(v) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.34875) = 0.559605 and Y_out(1.45125) = 0.0000430755. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v_line = 0.34875 + (0.00688605 − 0.559605)(1.45125 − 0.34875)/(0.0000430755 − 0.559605) = 1.43777 μeV. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives v = 1.02128 μeV; substituting back gives Y_out = 0.0068860479.
- The interpolation error in the input is |v_line − v| = 0.416492 μeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.0058596479 and 0.0079124479. Sort the roots even if the response decreases: 1.00692 μeV ≤ v ≤ 1.0377 μeV.
- The lower and upper input deviations are 0.0143524 μeV and 0.0164241 μeV. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer v ≈ 1.02128 μeV; compatible interval [1.00692, 1.0377] μeV. Endpoint interpolation gives 1.43777 μeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.